SPOJ DQUERY D-query（莫队基础题）

Posted 2020-10-25 leonard-

题目链接：http://www.spoj.com/problems/DQUERY/

题目：

Given a sequence of n numbers a1, a2, ..., an and a number of d-queries. A d-query is a pair (i, j) (1 ≤ i ≤ j ≤ n). For each d-query (i, j), you have to return the number of distinct elements in the subsequence ai, ai+1, ..., aj.

Input

• Line 1: n (1 ≤ n ≤ 30000).
• Line 2: n numbers a1, a2, ..., an (1 ≤ ai ≤ 106).
• Line 3: q (1 ≤ q ≤ 200000), the number of d-queries.
• In the next q lines, each line contains 2 numbers i, j representing a d-query (1 ≤ i ≤ j ≤ n).

Output

• For each d-query (i, j), print the number of distinct elements in the subsequence ai, ai+1, ..., aj in a single line.

   

Example

Input
5
1 1 2 1 3
3
1 5
2 4
3 5

Output
3
2
3 
题意：给定一个序列，询问m次，每次求出区间 [ L,R ] 不同数字出现的次数。
题解：莫队模板题

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 const int N=1e6+10;
 7 struct node{
 8     int l,r,id;
 9 }Q[N];
10 
11 int a[N],ans[N],cnt[N],BLOCK;
12 bool cmp(node x,node y){
13     if(x.l/BLOCK==y.l/BLOCK) return x.r<y.r;
14     return x.l/BLOCK<y.l/BLOCK;
15 }
16 
17 int n,m,Ans=0;
18 
19 void add(int x){
20     cnt[a[x]]++;
21     if(cnt[a[x]]==1) Ans++;
22 }
23 
24 void del(int x){
25     cnt[a[x]]--;
26     if(cnt[a[x]]==0) Ans--;
27 }
28 
29 int main(){
30     scanf("%d",&n);
31     BLOCK=sqrt(n);
32     for(int i=1;i<=n;i++){
33         scanf("%d",&a[i]);
34     }
35     scanf("%d",&m);
36     for(int i=1;i<=m;i++){
37         scanf("%d%d",&Q[i].l,&Q[i].r);
38         Q[i].id=i;
39     }
40     sort(Q+1,Q+1+m,cmp);
41     int L=1,R=0;
42     for(int i=1;i<=m;i++){
43         while(L<Q[i].l){
44             del(L);
45             L++;
46         }
47         while(L>Q[i].l){
48             L--;
49             add(L);
50         }
51         while(R<Q[i].r){
52             R++;
53             add(R);
54         }
55         while(R>Q[i].r){
56             del(R);
57             R--;
58         }
59         ans[Q[i].id]=Ans;
60     }
61     for(int i=1;i<=m;i++)
62     printf("%d\n",ans[i]);
63     return 0;
64 }